vibration chapter05
TRANSCRIPT
Mechanical Vibrations
Fifth Edition in SI UnitsSingiresu S. Rao
© 2011 Mechanical Vibrations Fifth Edition in SI Units3
Chapter 5Two Degree Freedom Systems
5
© 2011 Mechanical Vibrations Fifth Edition in SI Units4
Chapter Outline
5.1 Introduction5.2 Equations of Motion for Forced Vibration5.3 Free Vibration Analysis of an Undamped System5.4 Torsional System5.5 Coordinate Coupling and Principal Coordinates5.6 Forced-Vibration Analysis5.7 Semidefinite Systems 5.8 Self-Excitation and Stability Analysis5.9 Transfer-Function Approach5.10 Solutions Using Laplace Transform5.11 Solutions Using Frequency Transfer Functions
© 2011 Mechanical Vibrations Fifth Edition in SI Units5
5.1Introduction
5.1
© 2011 Mechanical Vibrations Fifth Edition in SI Units6
5.1 Introduction
• Two-degree-of-freedom systems are defined as systems that require two independent coordinates to describe their motion.
• For simplicity, a two-degree-of-freedom model can be used as shown in the figure.
© 2011 Mechanical Vibrations Fifth Edition in SI Units7
5.1 Introduction
• The general rule for the computation of the number of degrees of freedom can be stated as follows:
=
No. of degrees of freedom of the system
No. of masses in the system x no. of
possible types of motion of each mass
© 2011 Mechanical Vibrations Fifth Edition in SI Units8
5.1 Introduction
• As is evident from the systems shown in figures earlier on, the configuration of a system can be specified by a set of independent coordinates termed as generalized coordinates, such as length, angle, or some other physical parameters.
• Principle coordinates is defined as any set of coordinates that leads a coupled equation of motion to an uncoupled system of equations.
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5.2Equations of Motion for Forced Vibration
5.2
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5.2 Equations of Motion for Forced Vibration
• Consider a viscously damped two-degree-of-freedom spring-mass system, shown in the figure below
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5.2 Equations of Motion for Forced Vibration
• The application of Newton’s second law of motion to each of the masses gives the equations of motion:
• Both equations can be written in matrix form as
where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by
)2.5()()(
)1.5()()(
2232122321222
1221212212111
Fxkkxkxccxcxm
Fxkxkkxcxccxm
)3.5( )()(][)(][)(][ tFtxktxctxm
© 2011 Mechanical Vibrations Fifth Edition in SI Units12
5.2 Equations of Motion for Forced Vibration
• We have
• And the displacement and force vectors are given respectively:
• It can be seen that the matrices [m], [c], and [k] are symmetric:
where the superscript T denotes the transpose of the matrix.
322
221
322
221
2
1
][
][
0
0 ][
kkk
kkkk
ccc
cccc
m
mm
)(
)()(
)(
)()(
2
1
2
1
tF
tFtF
tx
txtx
][][],[][],[][ kkccmm TTT
© 2011 Mechanical Vibrations Fifth Edition in SI Units13
5.3Free-Vibration Analysis of an Undamped System
5.3
© 2011 Mechanical Vibrations Fifth Edition in SI Units14
5.3 Free-Vibration Analysis of an Undamped System
• The solution of Eqs.(5.1) and (5.2) involves four constants of integration (two for each equation). We shall first consider the free vibration solution of Eqs.(5.1) and (5.2).
• By setting F1(t) = F2(t) = 0, and damping disregarded, i.e., c1 = c2 =
c3 = 0, and the equation of motion is reduced to:
• Assuming that it is possible to have harmonic motion of m1 and m2
at the same frequency ω and the same phase angle Φ, we take the solutions as
)5.5(0)()()()(
)4.5(0)()()()(
2321222
2212111
txkktxktxm
txktxkktxm
)6.5()cos()(
)cos()(
22
11
tXtx
tXtx
© 2011 Mechanical Vibrations Fifth Edition in SI Units15
5.3 Free-Vibration Analysis of an Undamped System
• Substituting into Eqs.(5.4) and (5.5),
• Since Eq.(5.7)must be satisfied for all values of the time t, the terms between brackets must be zero. Thus,
which represent two simultaneous homogenous algebraic equations
in the unknown X1 and X2.
)7.5( 0)cos()(
0)cos()(
2322
212
221212
1
tXkkmXk
tXkXkkm
)8.5(0)(
0)(
2322
212
221212
1
XkkmXk
XkXkkm
© 2011 Mechanical Vibrations Fifth Edition in SI Units16
5.3 Free-Vibration Analysis of an Undamped System
• For trivial solution, i.e., X1 = X2 = 0, there is no solution. For a
nontrivial solution, the determinant of the coefficients of X1 and X2
must be zero:
which is called the frequency or characteristic equation.
0
)(
)(det
212
12
2212
1
kkmk
kkkm
)9.5(0))((
)()()(223221
1322214
21
kkkkk
mkkmkkmm
© 2011 Mechanical Vibrations Fifth Edition in SI Units17
5.3 Free-Vibration Analysis of an Undamped System
• Hence the roots are:
• The roots are called natural frequencies of the system.
)10.5())((
4
)()(
2
1
)()(
2
1,
2/1
21
223221
2
21
132221
21
13222122
21
mm
kkkkk
mm
mkkmkk
mm
mkkmkk
© 2011 Mechanical Vibrations Fifth Edition in SI Units18
5.3 Free-Vibration Analysis of an Undamped System
• To determine the values of X1 and X2,
• The normal modes of vibration corresponding to ω12 and ω2
2 can be
expressed, respectively, as
)11.5()(
)(
)(
)(
32222
2
2
21221
)2(1
)2(2
2
32212
2
2
21211
)1(1
)1(2
1
kkm
k
k
kkm
X
Xr
kkm
k
k
kkm
X
Xr
)12.5( and )2(
12
)2(1
)2(2
)2(1)2(
)1(11
)1(1
)1(2
)1(1)1(
Xr
X
X
XX
Xr
X
X
XX
© 2011 Mechanical Vibrations Fifth Edition in SI Units19
5.3 Free-Vibration Analysis of an Undamped System
• The free-vibration solution or the motion in time can be expressed itself as
• Initial conditions
The initial conditions are
(5.17)mode second)cos(
)cos(
)(
)()(
modefirst )cos(
)cos(
)(
)()(
22)2(
12
22)2(
1
)2(2
)2(1)2(
11)1(
11
11)1(
1
)1(2
)1(1)1(
tXr
tX
tx
txtx
tXr
tX
tx
txtx
0)0(,)0(
,0)0(constant, some )0(
2)(
12
1)(
11
txXrtx
txXtxi
i
i
© 2011 Mechanical Vibrations Fifth Edition in SI Units20
5.3 Free-Vibration Analysis of an Undamped System
• The resulting motion can be obtained by a linear superposition of the two normal modes, Eq.(5.13)
• Thus the components of the vector can be expressed as
• The unknown constants can be determined from the initial conditions:
)14.5()()()( 2211 txctxctx
)15.5()cos()cos(
)()()(
)cos()cos()()()(
22)2(
1211)1(
11
)2(2
)1(22
22)2(
111)1(
1)2(
1)1(
11
tXrtXr
txtxtx
tXtXtxtxtx
)16.5()0()0(),0()0(
),0()0(),0()0(
2222
1111
xtxxtx
xtxxtx
© 2011 Mechanical Vibrations Fifth Edition in SI Units21
5.3 Free-Vibration Analysis of an Undamped System
• Substituting into Eq.(5.15) leads to
• The solution can be expressed as
)17.5(sinsin)0(
coscos)0(
sinsin)0(
coscos)0(
2)2(
1221)1(
1112
2)2(
121)1(
112
2)2(
121)1(
111
2)2(
11)1(
11
XrXrx
XrXrx
XXx
XXx
)(
)0()0(sin,
)(
)0()0(sin
)0()0(cos,
)0()0(cos
122
2112
)2(1
121
2121
)1(1
12
2112
)2(1
12
2121
)1(1
rr
xxrX
rr
xxrX
rr
xxrX
rr
xxrX
© 2011 Mechanical Vibrations Fifth Edition in SI Units22
5.3 Free-Vibration Analysis of an Undamped System
• We can obtain the desired solution as
)18.5()0()0([
)0()0(tan
cos
sintan
)0()0([
)0()0(tan
cos
sintan
)0()0()0()0(
)(
1
sincos
)0()0()0()0(
)(
1
sincos
2112
2111
2)2(
1
2)2(
112
2121
2121
1)1(
1
1)1(
111
2/1
22
22112
21112
2/12
2)2(
1
2
2)2(
1)2(
1
2/1
21
22122
21212
2/12
1)1(
1
2
1)1(
1)1(
1
xxr
xxr
X
X
xxr
xxr
X
X
xxrxxr
rr
XXX
xxrxxr
rr
XXX
© 2011 Mechanical Vibrations Fifth Edition in SI Units23
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two-Degree-of-Freedom System
Find the free-vibration response of the system shown in Fig.5.3(a) with
k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1 and c1 = c2 = c3 = 0 for the
initial conditions .
© 2011 Mechanical Vibrations Fifth Edition in SI Units24
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two Degree of Freedom System
Solution
For the given data, the eigenvalue problem, Eq.(5.8), becomes
By setting the determinant of the coefficient matrix in Eq.(E.1) to zero, we obtain the frequency equation,
(E.1)0
0
55-
5 3510
0
0
2
1
2
2
2
1
322
22
2212
1
X
X
X
X
kkmk
kkkm
(E.2)01508510 24
© 2011 Mechanical Vibrations Fifth Edition in SI Units25
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two-Degree-of-Freedom System
Solution
The natural frequencies can be found as
The normal modes (or eigenvectors) are given by
E.3)(4495.2,5811.1
0.6,5.2
21
22
21
E.5)(5
1
E.4)(2
1
)2(1)2(
2
)2(1)2(
)1(1)1(
2
)1(1)1(
XX
XX
XX
XX
© 2011 Mechanical Vibrations Fifth Edition in SI Units26
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two-Degree-of-Freedom System
Solution
The free-vibration responses of the masses m1 and m2 are given by
(see Eq.5.15):
By using the given initial conditions in Eqs.(E.6) and (E.7), we obtain
(E.7))4495.2cos(5)5811.1cos(2)(
(E.6))4495.2cos()5811.1cos()(
2)2(
11)1(
12
2)2(
11)1(
11
tXtXtx
tXtXtx
(E.11)sin2475.121622.3)0(
(E.10)sin4495.2sin5811.10)0(
(E.9)cos5cos20)0(
(E.8)coscos1)0(
2)2(
1)1(
12
2)2(
11)1(
11
2)2(
11)1(
12
2)2(
11)1(
11
XXtx
XXtx
XXtx
XXtx
© 2011 Mechanical Vibrations Fifth Edition in SI Units27
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3Free-Vibration Response of a Two-Degree-of-Freedom System Solution
The solution of Eqs.(E.8) and (E.9) yields
The solution of Eqs.(E.10) and (E.11) leads to
Equations (E.12) and (E.13) gives
(E.12)7
2cos;
7
5cos 2
)2(11
)1(1 XX
(E.13)0sin,0sin 2)2(
11)1(
1 XX
(E.14)0,0,7
2,
7
521
)2(1
)1(1 XX
© 2011 Mechanical Vibrations Fifth Edition in SI Units28
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3Free-Vibration Response of a Two-Degree-of-Freedom System Solution
Thus the free vibration responses of m1 and m2 are given by
(E.16)4495.2cos7
105811.1cos
7
10)(
(E.15)4495.2cos7
25811.1cos
7
5)(
2
1
tttx
tttx
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5.4Torsional System
5.4
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5.4 Torsional System
• Consider a torsional system as shown in Fig.5.6. The differential equations of rotational motion for the discs can be derived as
• Upon rearrangement become
• For the free vibration analysis of the system, Eq.(5.19) reduces to
22312222
11221111
)(
)(
ttt
ttt
MkkJ
MkkJ
)19.5()(
)(
22321222
12212111
tttt
tttt
MkkkJ
MkkkJ
)20.5(0)(
0)(
2321222
2212111
ttt
ttt
kkkJ
kkkJ
© 2011 Mechanical Vibrations Fifth Edition in SI Units31
5.4 Torsional System
Example 5.4Natural Frequencies of a Torsional System
Find the natural frequencies and mode shapes for the torsional system
shown in the figure below for J1 = J0 , J2 = 2J0 and kt1 = kt2 = kt .
© 2011 Mechanical Vibrations Fifth Edition in SI Units32
5.4 Torsional System
Example 5.4Natural Frequencies of a Torsional SystemSolution
The differential equations of motion, Eq.(5.20), reduce to (with kt3 = 0,
kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):
Rearranging and substituting the harmonic solution:
(E.1) 02
02
2120
2110
tt
tt
kkJ
kkJ
(E.2)2,1);cos()( itt ii
© 2011 Mechanical Vibrations Fifth Edition in SI Units33
5.4 Torsional System
Example 5.4Natural Frequencies of a Torsional SystemSolution
This gives the frequency equation of
The solution of Eq.(E.3) gives the natural frequencies
(E.3)052 20
220
4 tt kkJJ
(E.4))175(4
and)175(4 0
20
1 J
k
J
k tt
© 2011 Mechanical Vibrations Fifth Edition in SI Units34
5.4 Torsional System
Example 5.4Natural Frequencies of a Torsional SystemSolution
The amplitude ratios are given by
Equations (E.4) and (E.5) can also be obtained by substituting the following in Eqs.(5.10) and (5.11).
(E.5)4
)175(2
4
)175(2
)2(1
)2(2
2
)1(1
)1(2
1
r
r
0and2,
,,
3022011
2211
kJJmJJm
kkkkkk tttt
© 2011 Mechanical Vibrations Fifth Edition in SI Units35
5.5Coordinate Coupling and Principal Coordinates
5.5
© 2011 Mechanical Vibrations Fifth Edition in SI Units36
5.5 Coordinate Coupling and Principal Coordinates
• Generalized coordinates are sets of n coordinates used to describe the configuration of the system.
• Equations of motion Using x(t) and θ(t)
© 2011 Mechanical Vibrations Fifth Edition in SI Units37
5.5 Coordinate Coupling and Principal Coordinates
From the free-body diagram shown in Figure (a), with the positive values of the motion variables as indicated, the force equilibrium equation in the vertical direction can be written as
The moment equation about C.G. can be expressed as
Eqs.(5.21) and (5.22) can be rearranged and written in matrix form as
)21.5()()( 2211 lxklxkxm
)22.5()()( 2221110 llxkllxkJ
)23.5(0
0
)( )(
)( )(
0
0 2
22
12211
221121
0 21
x
lklklklk
lklkkkx
J
m
© 2011 Mechanical Vibrations Fifth Edition in SI Units38
5.5 Coordinate Coupling and Principal Coordinates
The lathe rotates in the vertical plane and has vertical motion as
well, unless k1l1 = k2l2. This is known as elastic or static coupling.
•Equations of motion Using y(t) and θ(t)
From Figure b, the equations of motion for translation and rotation can be written as
melyklykym )()( 2211
)24.5()()( 222111 ymellykllykJP
© 2011 Mechanical Vibrations Fifth Edition in SI Units39
5.5 Coordinate Coupling and Principal Coordinates
These equations can be rearranged and written in matrix form as
If , the system will have dynamic or inertia coupling only.
•Note the following characteristics of these systems:
1.In the most general case, a viscously damped two degree of freedom system has the equations of motions in the form:
)25.5(0
0
)()(
)()(
2
22
112211
112221
2
y
lklklklk
lklkkky
Jme
mem
P
2211 lklk
)26.5(0
0
2
1
2221
1211
2
1
2221
1211
2
1
2221
1211
x
x
kk
kk
x
x
cc
cc
x
x
mm
mm
© 2011 Mechanical Vibrations Fifth Edition in SI Units40
5.5 Coordinate Coupling and Principal Coordinates
2. The system vibrates in its own natural way regardless of the coordinates used. The choice of the coordinates is a mere convenience.
3. Principal or natural coordinates are defined as system of coordinates which give equations of motion that are uncoupled both statically and dynamically.
© 2011 Mechanical Vibrations Fifth Edition in SI Units41
© 2011 Mechanical Vibrations Fifth Edition in SI Units42
5.5 Coordinate Coupling and Principal Coordinates
Example 5.6Principal Coordinates of Spring-Mass System
Determine the principal coordinates for the spring-mass system shown in the figure.
© 2011 Mechanical Vibrations Fifth Edition in SI Units43
5.5 Coordinate Coupling and Principal Coordinates
Example 5.6Principal Coordinates of Spring-Mass SystemSolution
Define two independent solutions as principal coordinates and express
them in terms of the solutions x1(t) and x2(t).
The general motion of the system shown is
(E.1)3
coscos)(
3coscos)(
22112
22111
tm
kBt
m
kBtx
tm
kBt
m
kBtx
© 2011 Mechanical Vibrations Fifth Edition in SI Units44
5.5 Coordinate Coupling and Principal Coordinates
Example 5.6Principal Coordinates of Spring-Mass SystemSolution
We define a new set of coordinates such that
Since the coordinates are harmonic functions, their corresponding equations of motion can be written as
(E.2)3
cos)(
cos)(
222
111
tm
kBtq
tm
kBtq
(E.3)03
and 0 2211
q
m
kqq
m
kq
© 2011 Mechanical Vibrations Fifth Edition in SI Units45
5.5 Coordinate Coupling and Principal Coordinates
Example 5.6Principal Coordinates of Spring-Mass SystemSolution
From Eqs.(E.1) and (E.2), we can write
The solution of Eqs.(E.4) gives the principal coordinates:
(E.4))()()(
)()()(
212
211
tqtqtx
tqtqtx
(E.5))]()([2
1)(
)]()([2
1)(
212
211
txtxtq
txtxtq
© 2011 Mechanical Vibrations Fifth Edition in SI Units46
5.6Forced-Vibration Analysis
5.6
© 2011 Mechanical Vibrations Fifth Edition in SI Units47
5.6 Forced-Vibration Analysis
• The equations of motion of a general two-degree-of-freedom system under external forces can be written as
• Consider the external forces to be harmonic:
where ω is the forcing frequency.
• We can write the steady-state solutions as
)27.5(
2
1
2
1
2221
1211
2
1
2221
1211
2
1
2212
1211
F
F
x
x
kk
kk
x
x
cc
cc
x
x
mm
mm
)28.5(2,1,)( 0 jeFtF tijj
)29.5(2,1,)( jeXtx tijj
© 2011 Mechanical Vibrations Fifth Edition in SI Units48
5.6 Forced-Vibration Analysis
• We can write Eq.(5.30) as:
where
• Eq.(5.32) can be solved to obtain:
)32.5()( 0FXiZ
20
100
2
1
2212
1211 matrix Impedance)( )(
)( )()(
F
FF
X
XX
iZiZ
iZiZiZ
)33.5()( 01FiZX
© 2011 Mechanical Vibrations Fifth Edition in SI Units49
5.6 Forced-Vibration Analysis
• The inverse of the impedance matrix is given
• Eqs.(5.33) and (5.34) lead to the solution
)34.5()( )(
)( )(
)()()(
1)(
1112
1222
2122211
1
iZiZ
i-ZiZ
iZiZiZiZ
)35.5()()()(
)()()(
)()()(
)()()(
2122211
201110122
2122211
201210221
iZiZiZ
FiZFiZiX
iZiZiZ
FiZFiZiX
© 2011 Mechanical Vibrations Fifth Edition in SI Units50
5.6 Forced-Vibration Analysis
Example 5.8
Steady-State Response of Spring-Mass System
Find the steady-state response of system shown in Fig.5.15 when the
mass m1 is excited by the force F1(t) = F10 cos ωt. Also, plot its
frequency response curve.
© 2011 Mechanical Vibrations Fifth Edition in SI Units51
5.6 Forced-Vibration Analysis
Example 5.8
Steady-State Response of Spring-Mass System
Solution
The equations of motion of the system can be expressed as
We assume the solution to be as follows
Eq.(5.31) gives
(E.1)0
cos
2
2
0
0 10
2
1
2
1
tF
x
x
k-k
-kk
x
x
m
m
E.2)(2,1;cos)( jtXtx jj
(E.3))(,2)()( 122
2211 kZkmZZ
© 2011 Mechanical Vibrations Fifth Edition in SI Units52
5.6 Forced-Vibration Analysis
Example 5.8
Steady-State Response of Spring-Mass System
Solution
Hence
Eqs.(E.4) and (E.5) can be expressed as
(E.5)))(3()2(
)(
(E.4)))(3(
)2(
)2(
)2()(
2210
22210
2
2210
2
22210
2
1
kmkm
kF
kkm
kFX
kmkm
Fkm
kkm
FkmX
E.6)(
1
2
)(2
1
2
1
2
1
2
10
2
1
1
k
F
X
© 2011 Mechanical Vibrations Fifth Edition in SI Units53
5.6 Forced-Vibration Analysis
Example 5.8
Steady-State Response of Spring-Mass System
Solution
E.7)(
1
)(2
1
2
1
2
1
2
102
k
FX
© 2011 Mechanical Vibrations Fifth Edition in SI Units54
5.7Semidefinite Systems
5.7
© 2011 Mechanical Vibrations Fifth Edition in SI Units55
5.7 Semidefinite Systems
• Semidefinite systems are also known as unrestrained or degenerate systems.
• Two examples of such systems are shown in the figure. For Figure (a), the equations of motion can be written as
• For free vibration, we assume the motion to be harmonic:
• Substituting Eq.(5.37) into Eq.(5.36) gives
)36.5(0)(
0)(
1222
2111
xxkxm
xxkxm
)37.5(2,1),cos()( jtXtx jjj
)38.5(0)(
0)(
22
21
212
1
XkmkX
kXXkm
© 2011 Mechanical Vibrations Fifth Edition in SI Units56
5.7 Semidefinite Systems
© 2011 Mechanical Vibrations Fifth Edition in SI Units57
5.7 Semidefinite Systems
• We obtain the frequency equation as
• From which the natural frequencies can be obtained:
• Such systems, which have one of the natural frequencies equal to zero, are called semidefinite systems.
)39.5(0)]([ 212
212 mmkmm
)40.5()(
and 021
2121 mm
mmk
© 2011 Mechanical Vibrations Fifth Edition in SI Units58
5.8Self-Excitation and Stability Analysis
5.8
© 2011 Mechanical Vibrations Fifth Edition in SI Units59
5.8 Self-Excitation and Stability Analysis
• Given that the criterion for stability is that the real parts of si must
be negative, all coefficients of equation ai must be positive and
hence, the condition
must be fulfilled.
• The Routh-Hurwitz criterion states that the system will be stable if
all the coefficients a0, a1,…,a4 are positive and the determinants
defined below are positive:
)46.5(214
230321 aaaaaaa
© 2011 Mechanical Vibrations Fifth Edition in SI Units60
5.8 Self-Excitation and Stability Analysis
)49.5(0
0
0
)48.5(0
)47.5(0
2304
21321
31
420
31
3
302120
312
11
aaaaaaa
aa
aaa
aa
T
aaaaaa
aaT
aT
© 2011 Mechanical Vibrations Fifth Edition in SI Units61
5.9Transfer-Function Approach
5.9
© 2011 Mechanical Vibrations Fifth Edition in SI Units62
5.9 Transfer-Function Approach
• For two-degree-of-freedom system shown the equations of motion are
• By taking the Laplace transforms of Eqs. (5.50) and (5.51), assuming zero initial conditions,
5.51
5.50
2122321223222
1221212212111
fxkxkkxcxccxm
fxkxkkxcxccxm
5.53
5.52
2122321223222
2
1221212212112
1
sFsXksXkkssXcssXccsXsm
sFsXksXkkssXcssXccsXsm
© 2011 Mechanical Vibrations Fifth Edition in SI Units63
5.9 Transfer-Function Approach
• It can be solved using Cramer’s rule as
where
5.57 and 5.56 22
11 sD
sDsX
sD
sDsX
© 2011 Mechanical Vibrations Fifth Edition in SI Units64
5.9 Transfer-Function Approach
• Note that
1. The denominator, D(s), in the expressions of X1(s) and X2(s)
given by Eq. (5.60), is a fourth-order polynomial in s and denotes the characteristic polynomial of the system The model (or system) is a fourth-order model (or system).
2. Equations (5.56) and (5.57) permit us to apply inverse Laplace transforms to obtain the fourth-order differential equations for
x1(t) and x2(t) .
3. Equations (5.56) and (5.57) can be used to derive the transfer
functions of and x2(t) corresponding to any specified forcing
function.
© 2011 Mechanical Vibrations Fifth Edition in SI Units65
5.10Solutions Using Laplace Transform
5.10
© 2011 Mechanical Vibrations Fifth Edition in SI Units66
5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Two railway cars, of masses m1 = M and m2 = m are connected by a
spring of stiffness k, as shown in the figure. If the car of mass M is subjected to an impulse determine the time responses of the cars using the Laplace transform method.
tF 0
© 2011 Mechanical Vibrations Fifth Edition in SI Units67
5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
The responses of the cars can be determined using either of the following approaches:a. Consider the system to be undergoing free vibration due to the initial velocity caused by the impulse applied to car M.b. Consider the system to be undergoing forced vibration due to the force applied to car M (with the displacements and velocities of cars M and m considered to be zero initially).
tF 0
© 2011 Mechanical Vibrations Fifth Edition in SI Units68
5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
Using the second approach, the equations of motion of the cars can be expressed as
Using the Laplace transforms, Eqs. (E.1) and (E.2) can be written as
E.2 0
E.1
122
0211
xxkxm
tFxxkxM
E.4 0
E.3
22
1
0212
sXkmsskX
FskXsXkMs
© 2011 Mechanical Vibrations Fifth Edition in SI Units69
5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
Equations (E.3) and (E.4) can be solved as
E.6
E.5
220
2
22
20
1
mMkMmss
kFsX
mMkMmss
kmsFsX
© 2011 Mechanical Vibrations Fifth Edition in SI Units70
5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
Using partial fractions, Eqs. (E.5) and (E.6) can be rewritten as
where
E.8 11
E.7 1
2220
2
2220
1
ws
w
wsmM
FsX
ws
w
wM
m
smM
FsX
E.9 112
mMkw
© 2011 Mechanical Vibrations Fifth Edition in SI Units71
5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
The inverse transforms of Eqs. (E.7) and (E.8), using the results of Appendix D, yield the time responses of the cars as
E.11 sin1
E.10 sin
02
01
wtw
tmM
FsX
wtwM
mt
mM
Fsx
© 2011 Mechanical Vibrations Fifth Edition in SI Units72
5.11Solutions Using Frequency Transfer Functions
5.11
© 2011 Mechanical Vibrations Fifth Edition in SI Units73
5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Derive the frequency transfer functions of x1(t) and x2(t) for the system
shown in figure.
© 2011 Mechanical Vibrations Fifth Edition in SI Units74
5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
From the free-body diagrams of the masses, the equations of motion of the system is
E.2 0
E.1 sin
212212222
01212212111111
pxxkxxcxm
wtPpxxkxxcxkxcxm
© 2011 Mechanical Vibrations Fifth Edition in SI Units75
5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
By taking the Laplace transforms of Eqs. (E.1) and (E.2), assuming zero initial conditions,
E.4 0
E.3
12212222
1212212111112
1
sXsXksXsXcsXm
sPsXsXksXsXcsXkssXcsXsm
© 2011 Mechanical Vibrations Fifth Edition in SI Units76
5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
The of Eqs. (E.3) and (E.4) is
where
E.6 and E.5 22
11 sD
sDsX
sD
sDsX
E.7
E.7
1222
1222
21
sPkscsD
sPkscsmsD
© 2011 Mechanical Vibrations Fifth Edition in SI Units77
5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
We have
The general transfer functions is
E.9 211221
221221221
3221221
422
kkskckc
scckmkmkmscmcmcmsmmsD
E.10 and E.9 22
1
2222
2
1
1
sD
ksc
sP
sX
sD
kscsm
sP
sX
© 2011 Mechanical Vibrations Fifth Edition in SI Units78
5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
The frequency transfer functions is
where
E.13 and E.12 22
1
2222
2
1
1
iwD
kiwc
iwP
iwX
iwD
kiwcwm
iwP
iwX
21122121221221
2
2212214
214
kkkckciwcckmkmkmw
cmcmcmiwwmmwiwD